Mr.Swanson drove 150 miles in 4 hours how many miles will he drive in 1 hour
1 answer:
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The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.
Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars (not all zero) such that
which leads to the system of linear equations,
From the third equation, we have , and substituting this into the second equation gives
and in turn, . Substituting these into the first equation gives
which tells us that any value of will work. If
, then
and
. Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.
Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.
Answer:
<h2>11.4</h2>Solution,
∆ ACB = ∆ EFD
finding the value of X,
Apply cross product property
Calculate the product
Divide both sides by 5
Calculate
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